I am trying to teach myself about p-adic numbers, and I came across an interesting property. Consider the following expansions:

In the case of , the repeating pattern is two digits, and similarly for negative two-thirds. In the case of the six expansions where the denominator is nine, the six digits shown repeat to the left. I understand multiplying by powers of 2 is essentially bit shifting, so those all make sense, but and . In other words, to get all rotations of the repeated digits, you simply multiply by all powers of 2 mod 9. This works when the denominator is 27 as well (there are 18 digits that repeat infinitely to the left, and 18 units of the local ring , and the operation of multiplication by 2 mod 27 does indeed rotate the 18 digits). Is there a way to prove this works when the denominator is any power of 3? I can prove that the number of repeated digits in the 2-adic expansion of is . This is actually relatively simple since a simple induction argument shows that always divides and

Then, since , which is a positive integer, its base-2 expansion takes at most digits, so the expansion repeats that many digits. Also, since multiplying by any power of 2 mod will give a unit of , multiplying the whole fraction by that unit will still take no more than digits to express in base-2, so that is still the number of repeating digits. I just don't have any idea how to show that you always get a rotation of the digits.

Edit: I am guessing the Division Algorithm might be useful. Consider the following:

has the solution . Now, where the six digits to the left of the five zeros adjacent to the decimal point are the repeated digits. Essentially, I just bit-shifted five places to the left. Now, which is also . I am not quite sure how to relate this to the rotation, though.

Aug 21st 2013, 09:16 AM

SlipEternal

Re: 2-adic expansions of inverse powers of 3

I think I might have figured it out. Would someone let me know if this proof suffices?

We know for any positive integer . Since , which is the inverse limit space, the division algorithm can be extended as follows. If the 2-adic expansion for , and I want , using the division algorithm, I would have

Since is a division ring, we know there exists such that . Plugging in for and multiplying both sides by , we get:

(Equation 1)

Since and are all integers, it must be that is an integer. Solving for , we get:

for any and .

Also,

So, we have . Next, if we divide both sides of (Equation 1) by , we get:

Since and , by the 3-adic extended division algorithm, we know that . So, from all of this, we obtain that . So, since , the multiplication by a unit of the local ring is essentially rotating the digits of the repeated pattern places to the right, which is exactly what I was trying to prove.

Edit: I am using the notation that is the space of 2-adic integers, not the set of congruence classes of , which is .

Aug 23rd 2013, 07:03 AM

SlipEternal

Re: 2-adic expansions of inverse powers of 3

A related question: does this property extend to higher primes? I think it does. Also if someone has a chance to check this out, can you let me know if this works, as well?

Let be primes. Then the p-adic expansion of has a repeating pattern (beginning at the decimal point) of digits. To show this, first I need to show that . I will prove this by induction. It is a simple algebraic fact that . Assume the property holds for all positive integers up to . Then by assumption, for some integer . Now, we have . Looking at the sum term-by-term, for all , is divisible by since for any . So, . Since , the property holds for all positive integers .

So,

Hence, the p-adic expansion of has repeating digits to the left of the decimal point as stated. So, assume you want to rotate the repeating pattern to the right places. We have ( for all ). Again, we want . Using the p-adic division algorithm, . Again, we know that is a division ring, so there exists such that . Plugging in for and multiplying both sides by gets the equation:

(Equation 1)

And again, we discover that is actually an integer (not just a p-adic integer). Then, dividing both sides by we get . So, . So, we would like so that the rotation of the digits places to the right corresponds to multiplying the original number by a unit of the ring . Solving for from Equation 1, we get which is greater than zero for any positive . Since , . So, as before, we have as desired, and is in fact a unit of the ring , which means so is .

Did I miss anything important? Are there any glaring holes in my argument?

Sep 4th 2013, 11:22 AM

SlipEternal

Re: 2-adic expansions of inverse powers of 3

I figured out where my proofs go wrong. I never prove that for every unit of the ring , there exists a number and a number such that . I still think it is true, but I need to figure out how to prove it in this direction.